As shown in the figure, let $P_1, P_2, P_3, P_4, P_5$ denote respectively the feet of perpendiculars from $P$ to segments $AB$, $BC$, $AC$, $BD$, $AD$.

Let $M_1, M_2, M_3, M_4, M_5$ be the midpoints of $AB$, $BC$, $AC$, $BD$, $AD$ respectively. Consider $\triangle ABC$. Clearly the circumcenter $O$ is the orthocenter of $\triangle M_1M_2M_3$. By Steiner's theorem, the reflections of line $L$ in $M_1M_2$, $M_2M_3$, $M_3M_1$ must meet at a point $K_1$ on the nine-point circle of $\triangle ABC$. Consider $\triangle ABD$, and determine the point $K_2$ by the same method. Since $K_1, K_2$ are both independent of the position of $P$, we will next show that $Q$ lies on the line $K_1K_2$, which will establish that the locus of $Q$ is a line.

We first prove: $\triangle POC \sim \triangle P_1 M_1 K_1$:
(i) Let the reflection of line $L$ in $M_2M_3$ be the line $EK_1$. Computing angles:
$$\begin{aligned} \angle K_1 M_1 B &= \angle EK_1 M_1 - \angle EFM_3 \\ &= \angle EM_2 M_1 - \angle PFM_3 \\ &= (\angle EM_2 M_3 + \angle M_3 M_2 M_1) - \angle PFM_3 \\ &= (\angle OCA + \angle CM_3 M_2) - \angle PFM_3 \\ &= \angle CGB - \angle PFM_3 = \angle COF. \end{aligned}$$
Hence we know $\angle P_1 M_1 K_1 = \angle POC$.
(ii) Computing the ratio:
$$\frac{M_1 K_1}{O C} = \sin \angle K_1 E M_1 = \sin \angle E O F = \frac{P_1 M_1}{P O}$$
(since the length $OC$ equals the diameter of the nine-point circle)
(iii) Hence by SAS we know $\triangle POC \sim \triangle P_1 M_1 K_1$. Similarly we also know: $\triangle POB \sim \triangle P_3 M_3 K_1$, $\triangle POA \sim \triangle P_2 M_2 K_1$.

Next we prove: $K_1$ lies on circle $c_1$; computing angles
$$\begin{aligned} \angle P_3 K_1 P_1 &= \angle M_3 K_1 M_1 - (\angle P_3 K_1 M_3 + \angle P_1 K_1 M_1) \\ &= \angle CAB - (\angle P_3 K_1 M_1 + \angle P_1 K_1 M_1) \quad (\text{since } K_1 \text{ lies on the nine-point circle}) \\ &= \angle CAB - (\angle PBO + \angle PCO) \\ &= (\angle OAC + \angle OBC) - (\angle PBO + \angle PCO) \\ &= (\angle OCA + \angle OCB) - (\angle PBO + \angle PCO) \\ &= \angle PCP_3 + \angle PBP_1 \\ &= \angle PP_2 P_3 + \angle PP_2 P_1 = \angle P_3 P_2 P_1, \end{aligned}$$
which proves that $K_1$ lies on circle $c_1$; similarly $K_2$ lies on circle $c_2$.
We next show that $\triangle P_1O_1O_2 \sim \triangle PCD$, where $O_1, O_2$ are respectively the centers of $c_1, c_2$:
(i) Let the points symmetric to $P$ with respect to $O_1, O_2$ be $I_1, I_2$ respectively. Then we know that $I_1, I_2$ are respectively the isogonal conjugate points of $P$ with respect to $\triangle ABC, \triangle ABD$. Computing
$$\begin{aligned} \angle AI_1B &= 180^\circ - (\angle I_1AB + \angle I_1BA) \\ &= 180^\circ - (\angle PAC + \angle PBC) \\ &= 180^\circ - (\angle APB - \angle ACB) \\ &= (180^\circ - \angle P_2P_1P_3). \end{aligned}$$
Similarly $\angle AI_2B = 180^\circ - (\angle APB - \angle ACB)$, hence $A, B, I_1, I_2$ are concyclic. Also we know: $\angle I_1BI_2 = \angle CBD$.
(ii)
$$\begin{aligned} \frac{O_1O_2}{CD} &= \frac{I_1I_2}{2CD} = \frac{I_1I_2}{AB} \times \frac{AB}{2CD} \\ &= \frac{1}{2} \frac{\sin \angle I_1BI_2}{\sin \angle AI_1B} \times \frac{\sin \angle ACB}{\sin \angle CBD} \\ &= \frac{1}{2} \frac{\sin \angle ACB}{\sin \angle AI_1B} \\ &= \frac{1}{2} \frac{\sin \angle ACB}{\sin \angle P_2P_1P_3} \times \frac{P_2P_3}{P_2P_3} \\ &= \frac{1}{2} \frac{2O_1P_1}{PC} = \frac{O_1P_1}{PC}. \end{aligned}$$
Similarly, $\frac{O_1O_2}{CD} = \frac{O_2P_2}{PD}$. Hence by SSS we obtain $\triangle P_1O_1O_2 \sim \triangle PCD$.

Since $\triangle POC \sim P_1M_1K_1$, $\triangle POD \sim \triangle P_1M_1K_2$, we therefore know that $\triangle PCD \sim \triangle P_1K_1K_2$. Hence
$$\triangle P_1 K_1 K_2 \sim \triangle P_1 O_1 O_2 \implies \angle P_1 O_1 K_1 = \angle P_1 O_2 K_2 \quad (\text{let this be } 2\alpha)$$
So we obtain: $\angle P_1QK_1 + \angle P_1QK_2 = \alpha + (180^\circ - \alpha) = 180^\circ$. So the three points $K_1, Q, K_2$ are collinear. But $K_1, K_2$ are both independent of the position of $P$, depending only on the line $L$, so as $P$ moves on $L$, the locus of the point $Q$ is a line $K_1K_2$.